Question
Abox of mass 60kg starts from the rest at height h and slides down a rough slope of length 10m, which makes an angle of 25 degree with the horizontal. It undergoes a constant accelequation of magnitude 2m/ while sliding down the slope. Calculate the work done on the box by the frictional force, using the work-energy theorem
Answers
Henry
Wb = m*g = 60kg * 9.8N/kg = 588 N. = Wt.
of box.
Fp = 588*sin25 = 248.5 N. = Force
parallel to the slope.
Fp-Fk = m*a
248.5-Fk = 60 * 2 = 120
Fk = 248.5 - 120 = 128.5 N. = Force of
kinetic friction.
Work = Fk * L = 128.5 * 10 = 1285 Joules
of box.
Fp = 588*sin25 = 248.5 N. = Force
parallel to the slope.
Fp-Fk = m*a
248.5-Fk = 60 * 2 = 120
Fk = 248.5 - 120 = 128.5 N. = Force of
kinetic friction.
Work = Fk * L = 128.5 * 10 = 1285 Joules
Lewis Kabika
Suppose you use the principle of work done by friction is equal to the change in the kinetic energy of the body. what would be the mistake?